Question

Answer the question below:

A rocket is launched from atop a 76-foot cliff with an initial velocity of 113 ft/s. The height of the rocket
above the ground at time is given by h = -161 +1131+ 76. When will the rocket hit the ground after it is
launched? Round to the nearest tenth of a second.
0.6 seconds
7.7 seconds
3.5 seconds
7.1 seconds

Answer 1

t=7.7 s

Explanation:

h=ut+1/2gt²

-76=113t+1/2(-32)t²

-76=113t-16t²

16t²-113t-76=0

`t=(113\pm√((-113)^2-4 * 16 *(-76)) )/(2 * 16) \\t=(113 \pm√(12769+4864) )/(32) \\t=(113 \pm√(17633) )/(32) \\t=(113+√(17633) )/(32) \approx 7.68~s \approx7.7 s\\or\\t=(113-√(17633) )/(32) \approx~-0.62 ~s \approx-0.6~s`

negative~sign~rejected.